The Specht problem in the variety of commutative
alternative algebras over a field of characteristic 3
Badeev A.V.
A variety of algebras is called satisfying the Specht property if every
of its subvarieties has a finite basis of identities. In 1984 S.V.~Pchelintsev
proved that every solvable alternative algebra $A$ over a field of
characteristic $\ne 2,3$ belongs to the variety
$\bf N_k N_2\bigcap N_3N_m$, i.e. for large enough $k,m$
$$(A^2)^k=(A^m)^3=0.$$
According to this result U.U.~Umirbaev proved that the variety of solvable
alternative algebras over a field of characteristic $\ne 2,3$ satisfies the
Specht property. The given result is the analogy of R.M.~Bryant's,
M.R.~Vaughan-Lee's and G.V.~Sheina's theorems for Lee algebras. The question
of the theorem justice was still open in the case of a field of
characteristic 3.
In the first part of this paper the following is proved:
\begin{theorem}
The variety $\bf N_kN_2\bigcap N_3N_m$ of commutative alternative algebras
over a field of characteristic 3 satisfies the Specht property.
\end{theorem}
In particular the variety of commutative alternative algebras defined by the
identity
$$[(x_1x_2)(x_3x_4)](x_5x_6)=0$$
satisfies the Specht property.
In the second part of this paper the Specht property is solved negatively in
case of a variety of solvable commutative alternative algebras. To be more
exact in the variety defined by the equation
$$[(x_1x_2)(x_3x_4)\cdot(x_5x_6)]x_7=0$$
the infinite system of identities of degree 4 in $x$:
$$[xR(x_1)\ldots R(x_{2n})\cdot xR(x_{2n+1})\ldots R(x_{4n})]\cdot
xR(x_{4n+1})\ldots R(x_{6n-1})R(x)=0$$
(where $R(x)$ is right multiplication operator) has not finite basis.